## Description

https://leetcode.com/problems/build-an-array-with-stack-operations/

Given an array `target`

and an integer `n`

. In each iteration, you will read a number from `list = {1,2,3..., n}`

.

Build the `target`

array using the following operations:

**Push**: Read a new element from the beginning`list`

, and push it in the array.**Pop**: delete the last element of the array.- If the target array is already built, stop reading more elements.

Return the operations to build the target array. You are guaranteed that the answer is unique.

**Example 1:**

Input:target = [1,3], n = 3Output:["Push","Push","Pop","Push"]Explanation:Read number 1 and automatically push in the array -> [1] Read number 2 and automatically push in the array then Pop it -> [1] Read number 3 and automatically push in the array -> [1,3]

**Example 2:**

Input:target = [1,2,3], n = 3Output:["Push","Push","Push"]

**Example 3:**

Input:target = [1,2], n = 4Output:["Push","Push"]Explanation:You only need to read the first 2 numbers and stop.

**Example 4:**

Input:target = [2,3,4], n = 4Output:["Push","Pop","Push","Push","Push"]

**Constraints:**

`1 <= target.length <= 100`

`1 <= target[i] <= n`

`1 <= n <= 100`

`target`

is strictly increasing.

## Explanation

Track the numbers added and compare with the target array to decide whether “push” or “pop” operation to use.

## Python Solution

```
class Solution:
def buildArray(self, target: List[int], n: int) -> List[str]:
results = []
numbers = []
for i in range(1, n + 1):
if numbers == target:
return results
numbers.append(i)
if numbers[len(numbers) - 1] == target[len(numbers) - 1]:
results.append("Push")
else:
results.append("Push")
results.append("Pop")
numbers.pop()
return results
```

- Time Complexity: O(N).
- Space Complexity: O(N).